Question

From a circular disc of radius $R$ and mass $9M$, a small disc of radius $R/3$ is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through $O$ is:

• A. $\frac{21}{5}M{R}^2$
• B. $\frac{40}{4}M{R}^2$
• C. $10M{R}^2$
• D. $\frac{37}{9}M{R}^2$

Answer 1

The correct option is A $\frac{21}{5}M{R}^2$

Initial moment of inertia of the disc $=\frac{1}{2}{\left(9M\right)R}^2$

$I=\frac{9}{2}{MR}^2$

Moment of inertia of the removed disc about their own centre and mass of the removed disc.

In ${\pi R}^2-9M$

In unit area $-\frac{9M}{{\pi R}^2}$

$\Rightarrow$ In ${\pi \left(\frac{R}{3}\right)}^2-\frac{9M}{{\pi R}^2}\times \pi {\left(\frac{R}{3}\right)}^2$

$m=M$

So,

$I_{removed\left(c\right)}=\frac{1}{2}M{\left(\frac{R}{3}\right)}^2$

Now, moment of inertia of removed portion about the centre of the bigger disc.

$I_{removed}=\frac{1}{18}{MR}^2+\frac{1}{2}M{\left(\frac{2R}{3}\right)}^2$

$I_{removeed}=\frac{{5MR}^2}{18}$

The net moment of inertia

$I_{finel}=\frac{9}{2}{MR}^2-\left(\frac{5}{18}\right){MR}^2$

$I_{finel}=\frac{21}{5}{MR}^2$